Solutions to Principles of Electronic Materials Part 2

* 1.9 Bonding and bulk modulus In general, the potential energy E per atom, or per ion pair, in a crystal as a function of interatomic (interionic) separation r can be written as the sum of an attractive PE and a repulsive PE, E(r) = − A B + rn rm General PE curve for bonding [1.49] where A and n are constants characterizing the attractive PE and B and m are constants characterizing the repulsive PE. This energy is minimum when the crystal is in equilibrium. The magnitude of the minimum energy and its location ro define the bonding energy and the equilibrium interatomic (or interionic) separation respectively. When a pressure P is applied to a solid, its original volume Vo shrinks to V by an amount ΔV = V − V0. The bulk modulus K relates the volume strain ΔV/V to the applied pressure P by P = −K(ΔV/Vo) Bulk modulus definition [1.50] The bulk modulus K is related to the energy curve. In its simplest form (assuming a simple cubic unit cell) K can be estimated from Equation 1.50 by K= 1 9cro d 2E   2 dr r =r o Bulk modulus [1.51] where c is a numerical factor, of the order of unity, given by b/p where p is the number of atoms or ion pairs in the unit cell and b is a numerical factor that relates the cubic unit cell lattice parameter ao to the equilibrium interatomic (interionic) separation ro by b = ao3 / ro3 a. Show that the bond energy and equilibrium separation are given by A n  E bond = n 1 −  r  m 1 and mBm− n ro =   nA  b. Show that the bulk modulus is given by K= An( m − n) + 9cron 3 K = Ebond or mn 9cro3 c. For a NaCl type crystal, Na+ and Cl− ions touch along cube edge so that ro = (ao/2). Thus, a3 = 23ro3 and b = 23 = 8. There are 4 ion pairs in the unit cell, p = 4. Thus, c = b/p = 8/4 = 2. Using the values from Example 1.3, calculate the bulk modulus of NaCl. Solution a. Interatomic separation r = ro is the distance at minimum E(r), Therefore we differentiate E(r) and set it equal to zero. i.e. dE (r )    = d  −A + B  = 0    dr r =r o dr  r n r m r = ro ∴ An mB   n+1 − m+1  =0 r r r = r o ∴ An mB =0 n +1 − m+1 ro ro ∴ rom +1 ro n +1 = mB nA rom −n = or mB nA 1 ∴ m −n mB  ro =   nA  (1) The potential energy is minimum at r = ro and is related with bonding energy E(ro) = −Ebond. From the equation for ro we have Bm m −n ro = An and isolate for B, B= Anrom− n m (2) Substitute for B in the energy relation − A B A 1 Anrom n   E ( ro ) = − n + m = − n + m    ro ro ro ro  m  m− n− m A Anr =− n + 0 ro m ∴ −n =− A Anr0 + ron m  A  n  E bond = −E (ro ) = −− n 1 −   ro  m =− A An + ron mron (3) ∴ A n  Ebond = n 1 −  ro  m b. Show that the bulk modulus is given by An K = n +3 (m − n ) or 9cro (4) K= mnE bond 3 9cro From the definition of Bulk modulus mentioned in the problem statement above 1 d 2 E  K =   9cr 0 dr 2 r =r0 First, we find d2E , i.e. dr 2 dE ( r ) = d  − A + B  An mB =  n − r m r n+1 r m+1 dr dr  r ∴ dE2 ( r)  d  An mB   2  =  n +1 − m +1   r r = r0 dr  r =ro dr r − n (n + 1)A m( m +1)B − n(n + 1)A m(m+ 1)B + = +  = + m +2 n+2  r n +2 r = r 0 ro m 2 r ro Again, substituting the value of B in the above relation, i.e. B = Anrom m −n we have dE 2 ( r )  − n( n +1) A m( m +1) Anr0m−n − n( n+1) A An( m +1) + = + m+2 − m+ n  2  = n +2 m+ 2 n +2 ro ro m ro  dr r = r o ro ∴ dE 2 (r )  − n(n + 1)A An(m + 1) An + = n+ 2 (− n − 1 + m + 1)  2  = n+2 n+2 ro ro ro  dr r = ro ∴ dE 2 ( r )  An  2  = n +2 (m − n ) dr  r =ro ro Now substitute for the second derivative in the equation for the Bulk modulus 2 An( m −n)  1 d E  1    K=  2 =  n+ 2  9cro dr r =r0 9cr o  r o  or K= An( m − n) + 9cron 3 From the relationship for the bonding energy, A  −n  A( m − n) = − E bond = − n  1 ro  m  ron m (5) Now, consider the expression for K and rearrange it as K= i.e. A(m − n ) mn ronm 9 cro3 K = E bond mn 9cro3 c. From Example 1.3, the bonding energy for NaCl is M = 1.748, n = 1, m = 8, ro = 0.281 × 10−9 m, c = 2. Therefore, 2 2 −19 e M (1.6 ×10 C) (1.748) A= = = 4.022 × 10−28. − 12 4πε o 4π (8.85× 10 F m− 1 ) Substitute the above value for A in the expression for K in Equation (5), An(m − n ) (4.022× 10−28 )(1)(8− 1) K= = 25.1 × 109 Pa or 25.1 GPa = 9 ⋅(2)(0.281 ×10− 9 )1+ 3 9 cron +3 -----------------------Note: Experimental value is roughly 2.4×1010 Pa or 24 GPa. The calculated value is quite close. (L.M. Thomas and J. Shanker, "Equation of State and Bulk Modulus for NaCl", Physica Status Solidi B, 189, 363 (1995)) Comment: Equation (3) written as A n  E (r o ) = − n 1 −  ro  m is often called Born-Landé equation for the lattice energy of a crystal (after Max Born and Alfred Landé) 1.10 Van der Waals bonding Below 24.5 K, Ne is a crystalline solid with an FCC structure. The interatomic interaction energy per atom can be written as 6 12  σ  σ   E ( r ) = −2ε 14.45  − 12.13  (eV/atom) r  r      where ε and σ are constants that depend on the polarizability, the mean dipole moment, and the extent of overlap of core electrons. For crystalline Ne, ε = 3.121 × 10−3 eV and σ = 0.274 nm. a. Show that the equilibrium separation between the atoms in an inert gas crystal is given by ro = (1.090)σ . What is the equilibrium interatomic separation in the Ne crystal? b. Find the bonding energy per atom in solid Ne. c. Calculate the density of solid Ne (atomic mass = 20.18). Solution a. Let E = potential energy and x = distance variable between the atoms. The energy E is given by 6 12  σ  σ   E( x) = −2ε  14.45   −12.13  x  x      The force F on each atom is given by 11 5  σ  σ    σ  σ   dE ( x) x  x   F ( x) = − 145.56 = 2ε  − 86.7 2  x2  dx x       ∴  σ 12 σ6 F ( x) =2ε 145 .56 13 −86.7 7 x x   When the atoms are in equilibrium, this net force must be zero. Using ro to denote equilibrium separation, F (ro ) = 0 ∴  σ 12 − σ6 2 ε 145 .7 7 =0  . 56 13 86  ro ro    ∴ 145.56 13 ∴ ∴ ro ro 7 σ 12 ro 13 = 86.7 σ6 ro7 12 145.56 σ  =  86.7 σ6 ro = 1.090 σ For the Ne crystal, σ = 2.74 × 10−10 m and ε = 0.003121 eV. Therefore, ro = 1.090(2.74 × 10−10 m) = 2.99 × 10−10 m for Ne. b. Calculate energy per atom at equilibrium:  σ E( ro ) = −2 ε  14.45 r  o  ∴ ∴ 6  σ   − 12.13 r  o (      12 − E ( ro ) = −2(0.003121 eV ) 1.602 ×10 19 6   2.74 ×10 -10 m  14.45     -10   2.99 × 10 m  J/eV  12  -10 2.74 × 10 m     −12.13    -10  2.99× 10 m    ) E(ro) = −4.30 × 10−21 or −0.0269 eV Therefore the bonding energy in solid Ne is 0.027 eV per atom or 2.6 kJ / mole c. To calculate the density, remember that the unit cell is FCC, and density = (mass of atoms in the unit cell) / (volume of unit cell). There are 4 atoms per FCC unit cell, and the atomic mass of Ne is 20.18 g/mol. (See Figure 1Q10-1) Figure 1Q10-1: Left: An FCC unit cell with close-packed spheres. Right: Reduced-sphere representation of the FCC unit cell. Since this is an FCC crystal structure, let a = lattice parameter (side of cubic cell) and R = radius of atom. The shortest interatomic separation is ro = 2R (atoms in contact means nucleus to nucleus separation is 2R (see Figure 1Q10-1). R = ro/2 and 2a 2 = (4R)2 ∴ r  a = 2 2R = 2 2 o = 2 2.99 × 10 −10 m 2  ( ) a = 4.228 × 10 −10 m or 0.423 nm ∴ The mass (mNe) of 1 Ne atom in grams is the atomic mass (Mat) divided by NA (Avogadro's number) because NAnumber of atoms have a mass of Mat grams, mNe = Mat / NA mNe = ∴ ( 20.18 g/mol)( 0.001 kg/g) = 3.351×10− 26 kg 6.022×1023 mol-1 There are 4 atoms per unit cell in the FCC lattice. The density ρ can then be found by ρ = (4mNe ) / Volume of unit cell ρ = (4mNe ) / a3 = [4 × (3.351 × 10−26 kg)] / (4.228 × 10−10 m) 3 or ρ = 1774 kg/m 3 ∴ 3 In g/cm , this density is: 3 ρ= 1774 kg/m × (1000 g/kg ) = 1.77 g/cm3 (100 cm/m)3 The density of solid Ne is 1.77 g cm−3. ------------------------ 1. Except for He, all inert gas crystal structures are FCC. 2. The calculated value is very close to the experimental value for a (below 24 K, or −246 °C), which is 0.44 nm. (e.g. https://www.webelements.com/neon/crystal_structure.html, 22 October 2016). The values of ε and σ for Ne are from C. Kittel, Solid State Physics, Seventh Edition, John Wiley and Sons, New York, 1996; Table 4, p60. Values in SI units are given below He ε x 10-21 J ε (meV) σ (nm) 0.14 Ne Ar Kr Xe 0.5 1.67 2.25 3.2 0.874 3.121 10.424 14.045 19.975 0.256 0.274 0.34 0.365 0.398 3. There is an excellent undergraduate level description and discussion of Lennard-Jones potential for two isolated atoms and atoms in the FCC crystal in Alan Walton, Three Phases of Matter, Second Edition, Clarendon Press (Oxford University Press), Oxford, 1983; pp43-47 for two isolated atoms and pp259-261 for atoms in the crystal. 1.11 Kinetic molecular theory a. In particular Ar-ion laser tube the gas pressure due to Ar atoms is about 0.1 torr at 25 °C when the laser is off. What is the concentration of Ar atoms per cm3 at 25 °C in this laser? (760 torr = 1 atm = 1.013×105 Pa.) b. In the He-Ne laser tube He and Ne gases are mixed and sealed. The total pressure P in the gas is given by contributions arising from He and Ne atoms P = PHe + PNe where PHe and PNe are the partial pressures of He and Ne in the gas mixture, that is, pressures due to He and Ne gasses alone, NHe RT  N Ne RT  and     PHe = PNe = N A V  NA V  In a particular He-Ne laser tube the ratio of He and Ne atoms is 7:1, and the total pressure is about 1 torr at 22 °C. Calculate the concentrations of He and Ne atoms in the gas at 22 °C. What is the pressure at an operating temperature of 130 °C? Solution a. From the Kinetic molecular theory for gases, we have N  PV =    RT N A  where, R is the gas constant constant, T is the temperature. The number of Ar atoms per unit volume is N PN A n Ar = = V RT We are given P = 0.1 torr × 1.013× 10 5 Pa = 13.33 Pa 760 torr Therefore the number of Ar atoms per unit volume nAr will be − n Ar = (13.33 Pa)(6.022 ×1023 mol 1 ) = 3.24 × 1021 m−3 or 3.24 × 1015 cm−3 −1 −1 (8.3145 J K mol )(25 + 273 K) b. Let nHe = NHe/V, the concentration of He atoms; nNe = NNe/V, the concentration of Ne atoms. Given that the total pressure is the sum of the pressure by He and Ne gasses P = PHe + PNe N He RT N Ne RT RT  RT     ∴ + = nHe  P =       + nNe    N V N V N       N  A ∴ A A A RT  RT  RT   P = 7 nNe    + n Ne    = 8nNe    N A  N A   NA  PN A 8RT Thus at T = 22 °C (295 K) for Ne, ∴ n Ne = (1 torr ) −1 23 (6.022 ×10 mol ) (760 torr ) 8(8.3145 J K −1mol −1 )(273 + 22 K ) 5 (1.1013 × 10 Pa ) n Ne = = 4.09 × 1021 m−3 or 4.09 × 1015 cm−3. Given that nHe is 7 times that of nNe, i.e. nHe = 7× nNe = 2.86× 1022 m−3 or 2.86× 1016 cm−3. At T = 130 °C (403 K), the atomic concentrations of He and Ne remain unchanged (the tube has the same volume, neglecting the thermal expansion). Thus, the new pressure P′ and initial pressure P are related by P′ RT′ / V T ′ (273 +130 K) = = = = 1.366 P RT/ V T ( 273 + 22 K) so that the new pressure P' is 1.37 torr. ____________________________________________________________________________________ 1.12 Kinetic Molecular Theory Calculate the effective (rms) speeds of the He and Ne atoms in the HeNe gas laser tube at room temperature (300 K). Solution Figure 1Q12-1: The gas atoms in the container are in random motion. Figure 1Q12-2: The He-Ne gas laser. To find the root mean square velocity (vrms) of He atoms at T = 300 K: The atomic mass of He is (from Periodic Table) Mat = 4.0 g/mol. Remember that 1 mole has a mass of Mat grams. Then, one He atom has a mass (m) in kg given by: m= Mat 4.0 g/mol × (0.001 kg/g ) = 6.642 ×10− 27 kg = − NA 6.022 × 10 23 mol 1 From the kinetic theory (visualized in Figure 1Q12-1), 1 3 2 m(v rms ) = kT 2 2 ∴ v rms = 3 ( ) kT 1.381× 10 J K ( 300 K) = 3 = 1368 m/s − m 6.642 ×10 27 kg − 23 ( -1 ) The root mean square velocity (vrms) of Ne atoms at T = 300 K can be found using the same method as above, changing the atomic mass to that of Ne, Mat = 20.18 g/mol. After calculations, the mass of one Ne atom is found to be 3.351 × 10−26 kg, and the root mean square velocity (vrms) of Ne is found to be vrms = 609 m/s. -----------------------Note: Radiation emerging from the He-Ne laser tube (Figure 1Q12-2) is due to the Ne atoms emitting light, all in phase with each other, as explained in Ch. 3. When a Ne atom happens to be moving towards the observer, due to the Doppler Effect, the frequency of the laser light is higher. If a Ne atom happens to moving away from the observer, the light frequency is lower. Thus, the random motions of the gas atoms cause the emitted radiation not to be at a single frequency but over a range of frequencies due to the Doppler Effect. ________________________________________________________________________________ *1.13 Kinetic molecular theory and the Ar ion laser An argon ion laser has a laser tube that contains Ar atoms that produce the laser emission when properly excited by an electrical discharge. Suppose that the gas temperature inside the tube is 1300 °C (very hot). a. Calculate the mean speed (vav), rms velocity (vrms = v 2 ) and the rms speed (vrms,x = v x2 ) in one particular direction of the Ar atoms in the laser tube, assuming 1300 °C. (See Example 1.11.) b. Consider a light source that is emitting waves and is moving towards an observer, somewhat like a whistling train moving towards a passenger. If fo is the frequency of the light waves emitted at the source, then, due to the Doppler effect, the observer measures a higher frequency f that depends on the velocity vAr of the source towards to observer and the speed c of light,  v  f =f o1 + Ar  c  It is the Ar ions that emit the laser output light in the Ar-ion laser. The emission wavelength λo = c/fo is 514.5 nm. Calculate the wavelength λ registered by an observer for those atoms that are moving with a mean speed vav toward the observer. Those atoms that are moving away from the observer will result in a lower observed frequency because vAr will be negative. Estimate the width of the wavelengths (the difference between the longest and shortest wavelengths) emitted by the Ar ion laser. Solution a. From Example 1.11 the mean speed is given by 8 kT v av = πm m is the mass of a gas atom. T = 1300 °C = 1573 K. Atomic mass of Ar is Mat = 39.95 g mol−1, therefore the mass of the Ar atom is − M 39.95 g mol 1 −23 m = at = g or 6.634 × 10−26 kg − = 6.634 × 10 23 N A 6.022 ×10 mol 1 Mean speed is given by v av = ∴ −23 −1 −23 −1 8kT 8(1.38066 ×10 JK )(1300 + 273 K ) = πm π (6.634 × 10 −26 kg) vav = 913.32 m s−1. Root mean square RMS velocity is v rms = ∴ 3kT = m 3(1.3806 ×10 J K )(1300 + 273 K ) − (6.634 × 10 26 kg) vrms = 991.31 m s−1. RMS speed alon x, vrms,x = v x2 v rms,x = v rms 3 = in one particular direction is 991.31m s 3 −1 = 572.33 m s−1. b. First we consider the case when the source is moving towards the observer with average speed, vav, the frequency observed is  v  f = f o1 + Ar  c  where fo = c/λ o = 3×108 m s−1/ 514.5 ×10−9 m = 5.8309 × 1014 s−1, vAr as calculated above is 912.32 m s−1. Therefore the frequency is  913.32 m s −1  = 5.830922× 1014 s−1 f1 = 5.8309 × 1014 s −1  1+ 8 −1   3× 10 m s  The corresponding wavelength is therefore, λ 1 = c/f1 = 3×108 m s−1/ 5.830922× 1014 s−1 = 514.4984 nm. In the case when the emitting source is moving away from the observer, the frequency is  913 .32 m s −1  − = 5.830886× 1014 s−1 f 2 = 5.8309 ×10 14 s 1  1 − 8 −1   3 × 10 m s  The corresponding wavelength is therefore, λ 2 = c/f2 = 3×108 m s−1/ 5.830886× 1014 s−1 = 514.5016 nm. The range of wavelengths observed by the observer is between 514.4984 nm and 514.5016 nm. The wavelength (spectral) width is Δλ= λ2 – λ1 = 514.5016 − 514.4984 nm = 0.0032 nm, very small. -----------------------Note: The question asks for the change in the wavelength or the width in the emitted wavelengths. Four decimal places were kept in the calculations of frequency and wavelength because we are interested in these changes and the changes in the frequency and wavelength are small. It may be thought that we should similarly use higher accuracy in the velocity calculations and a more accurate c value etc but that's not necessary because the change in the frequency is actually 2fovAr/c: v  v   v  Δf =f o 1 + Ar  −f o1 − Ar = 2 f o Ar  c   c  c = 2(5.8309 × 1014 s−1)(913.32 m s−1) / (3 × 108 m s−1) = 3.55 × 109 s−1 = 3.55 GHz. Note to the Instructor: Some students are known to convert a range of frequencies to a range of wavelengths by taking Δλ = c/Δf, which is wrong. To convert a small range of frequencies Δf to a range of wavelengths Δλ, take λ = c/f and differentiate it, dλ c =− 2 df f 8 ∴ Δλ ≈ −1 dλ c 3 × 10 m s Δf = 2 Δ f = (3.55 ×10 9s −1) = 3.13 × 10−12 m = 0.00313 nm df fo (5.8309 ×10 14 s −1) 2 very close to the above calculation of 0.0032 nm. 1.14 Heat capacity of gases Table 1.9 shows the experimental values of the molar heat capacity for a few gases at 25 °C. Assume that we can neglect the vibrations of the atoms in the molecules. For each calculate the observed degree of freedom fobserved, that is f in Cm = f(R/2). For each find the expected fexpected by considering translational and rotational degrees of freedom only. What is your conclusion? Table 1.9 Heat capacities for some gases at room temperature at constant volume, CV in J mol−1 K−1. N2 CO2 CH4 SF6 Gas Ar Ne Cl2 O2 CV 12.5 12.7 25.6 21.0 20.8 28.8 27.4 89.0 Solution If a gas atom or molecule has f degrees of freedom, then its molar heat capacity at constant volume is given by CV = f(R/2) = f(4.158 J mol−1 K−1). Thus we can obtain the experimental f as fobserved = CV ( measured) = 3.00 R which for the Ar gas gives fobserved = 12.5 = 3.00 4.158 This is exactly what we expect since the Ar gas consists of single Ar atoms moving around randomly in the tank. The Ar values have been entered into Table 1Q14-1. The remainder of the gases are similarly calculated and entered into Table 1Q14-1. First, notice that for monoatomic gases such as Ar and Ne, fobserved and fexpected agree very well. For diatomic molecules such as Cl2, O2, N2, we expect 5 degrees of freedom (DOF), 3 translational and 2 rotational, ignoring the vibrational degrees of freedom. O2 and N2 follows the expected behavior at room temperature; but not Cl2. The latter exhibits more DOF than 5 because the Cl2 molecules can gain energy through the vibrations of the two Cl atoms held together by a bond that acts as a spring (Cl−Cl). The vibrational DOF would normally contribute 2 additional (R/2) values but at room temperature their contribution is not full. (Full contribution means R/2 per DOF) CO2 is actually a linear molecule and should have 5 translational and rotational DOF without the vibrations. It shows nearly 7 DOF, arising from various vibraions. Both CO2 and CH4 should have full 3 translational and 3 rotational DOF, i.e. fexpected = 6. However, their DOF is more than 6. Again there is a contribution from vibrational DOF. SF6 should have 3 translational and 3 rotational degrees of freedom, 6 in total but it exhibits some 21 DOF. Clearly vibrations play a very important role in this gas Table 1Q14-1 Heat capacities for some gases at room temperature at constant volume, CV in J mol−1 K−1 and observed and expected f. Gas CV fobserved fexpected (translation + rotation only) Ar 12.5 3.01 Ne 12.7 3.05 Cl2 25.6 6.16 O2 21.0 5.05 N2 20.8 5.00 CO2 28.9 6.95 CH4 27.4 6.59 SF6 89.0 21.40 3 3 5 5 5 5 6 6 Note: Data in Table 1.9 on elements are from *1.15 Degrees of freedom in a gas molecule A monatomic molecule such as Ar has only three degrees of freedom (DOF) for motion along the three independent directions x, y, and z. In a system in which there are two independent atoms such Cl and Cl, the total number of DOF f is 6 because each atom has 3 degrees of freedom. Once we form a Cl2 molecule, the original 6 DOF in KE are partitioned as shown in Figure 1.78 The Cl2 molecule has 3 translational degrees of freedom, 2 rotational and 1 vibrational, summing to the original 6. The vibrational degree of freedom itself has KE and PE terms with each భ kT so that a vibrational degree of freedom actually has kT of energy rather than భkT. having an average of మ మ The PE term arises from the stretching and compression of the bond (which acts like a spring) during the vibrations. Put differently, each vibrational DOF has two subdegrees of freedom associated with KE and PE terms, each of which has (1/2)kT of energy. Let na be the number of atoms in a molecule. Then 3na is the total number of kinetic energy based DOF. There will always be 3 translational DOF for the molecule and at most 3 rotational degrees of freedom. There may be one or more vibrational DOF because there may be many ways in which the atoms in the molecule can vibrate, but there is a maximum. If frot and fvib are the rotational and vibrational DOF, then 3na = 3 + frot + fvib. a. What is the vibrational DOF for Cl2. What is the maximum molar heat capacity at constant volume CV for Cl2? Given Table 1.9, what is the vibrational contribution? b. What is the vibrational DOF for SF6. The molar heat capacity at constant volume for the SF6 gas at 300 K is 89.0 J mol−1 K−1 but at 700 K, it is 141 J mol−1 K−1. How many vibrational DOF do you need to explain the observations at these two temperatures? Figure 1.78 The partitioning of degrees of freedom in a diatomic molecule. Solution a. The Cl2 molecule has 2 atoms, joined by a spring (bond). There are 2 rotational DOF, we neglect the rotation about the bond (as shown in Figure 1.78). Thus, the general rule for the DOF, 3na = 3 + frot + fvib gives fvib = 3na − 3 − frot = 3(2) − 3 − 2 = 1 Thus, the Cl2 molecule has 1 vibrational DOF, which itself has 2 "subdegrees" of freedom arising from the KE and PE terms. The molar constant volume heat capacity CV at most should be CV = translational DOF × (R/2) + rotational DOF × (R/2) + vibrational DOF × R ∴ CV = 3(R/2) + 2(R/2) + 1R = (7/2)R = 29.1 J mol−1 K−1 Table 1.9 gives CV for Cl2 at 300 K as 25.6 J mol−1 K−1, not the expected full value of 29.1 J mol−1 K−1; less than the expected. Without the vibrations this CV would have been 5(R/2) or 20.79 J mol−1 K−1. The difference between the measured and 20.79 J mol−1 K−1 is the vibrational contribution to CV, that is Vibrational contribution to CV at 300 K = 25.6 − 20.79 = 4.81 J mol−1 K−1. The full vibrational contribution would have been 1R, that is 8.316 J mol−1 K−1. Clearly, the vibrational contribution is about half its full capacity. b. The SF6 molecule has 7 atoms. There are 3 rotational DOF. Thus, the general rule for the DOF, 3na = 3 + frot + fvib gives fvib = 3na − 3 − frot = 3(7) − 3 − 3 = 15 Thus, the SF6 molecule has 15 vibrational DOF, with each having 2 "subdegrees" of freedom arising from the KE and PE terms. The molar constant volume heat capacity CV at most should be CV = translational DOF × (R/2) + rotational DOF × (R/2) + vibrational DOF × R ∴ CV = 3(R/2) + 3(R/2) + 15R = 18R = 149.7 J mol−1 K−1 Consider now the CV-contributions from translational and rotational motions only (non-vibrational contribution), represented by CV(TR) CV(TR) = 3(R/2) + 3(R/2) = 25 J mol−1 K−1 The measured vibrational contribution CV(V) at 300 K is therefore CV(V) = CV(Measured) − CV (TR) = 89 J mol−1 K−1 − 25 J mol−1 K−1 = 64 J mol−1 K−1 It is clear that the vibrational contribution is very strong; stronger than the translational and rotational contribution. Consider Vibrational DOF = CV(V) / R = (64 J mol−1 K−1) / (8.3145 J mol−1 K−1) = 7.7 Recall that with the vibrational DOF, each DOF has an average energy of R, and be considered to be made up of KE and PE associated "subdegrees" of freedom; each with an energy R/2. Clearly at least half the vibrations are contributing to CV. Now consider CV at 700 K (427 °C), which is 141 J mol−1 K−1. Thus, the vibrational contribution to the molar heat capacity is